Solutions of Gateway Examinations to IISER, NISER, JEE, NEET, NDA, ICAR, OUAT
Solutions of Gateway Examinations to
IISER, NISER, JEE, NEET,
NDA, ICAR, OUAT
Topic: Solution (NDA, ICAR, OUAT Level)
1. Solubility of a gas in a solvent varies directly with the prssure of the gas, inversly with the temperature of the solution and directly with the liquefiability of a gas. So all of the above are true.
2. Calculate the amount of urea dissolved per litre if its aqueous solution is isotonic with 10% cane sugar solution.
Isotonic Solutions are those having equal osmotic pressure (O.P).
O.P = w R T / m V
In case of Urea, w1 = ?, m1 = 60, V1 = 1 litre
In case of Cane sugar, w2 = 10g, m2 = 342, V2 = 100ml = 0.1 litre
Thus w1 R T / m1 V1 = w2 R T / m2 V2
Or w1 / m1 V1 = w2 / m2 .V2
Thus w1 = 17.543 g = 17.6 g
3. If we prepare separate solutions of 2 mole urea and 2 moles NaCl in one litre water, then whose boiling point will be higher?
Number of particles in NaCl solution will be greater due to ionisation hence the value of any colligative properties will be higher in this case.
Urea is non electrolyte in water giving less number of particles.
4. Mole fraction of ethanol in ethanol - water mixture is 0.25. What is percent concentration of ethanol by weight of solution?
X (C2H5OH) = 0.25 which means X (H2O) = 0.75
When such data are given without total moles, we can consider any value of total moles, cause mole fraction is the moles of a component out of total moles.
Let us consider total moles to be 100. Thus the mixture will contain, 25 moles of ethanol and 75 moles of water. As we got the number of moles then we can calculate the mass of solute and solution and then applying m/m %age formula we can get the answer.
m/m %age = (25 X 46) / {25 X 46 + 75 X 18} = 46 %
5. Maximum freezing point will be for 1 molal solution of (assume equal ionisation in each case) [Fe(H2O)6]Cl3 , [Fe (H2O)5 Cl] Cl2.H2O , [Fe (H2O)4 Cl2] Cl. 2H2O , [Fe (H2O)3 . Cl3] 3H2O .
Greater is the number of particles formed, higher will be thevaue of colligative properties, that means lower will be the boiling point. The compound which gives smallest number of particles will have smaller colligative properties and maximum freezing point. The answer is, [Fe (H2O)3 . Cl3] 3H2O as it does not form ions and in total give one particle only.
6. Lowering of vapor pressure due to a solute in 1 molal aqueous solution at 100 degree centigade is:
A problem in which you are unable to find sufficient idea to solve, look into the data given.
Lowering of vapour means it must contain the vapour pressure of pure solvent. In this case the vapour pressure of pure solvent at 100 degree centigrade is 1 atm , because the at the boiling point the v.p becomes equal to the atmospheric pressure. Thus the V.P of pure solvent = p0 = 1atm = 760 mm 760 torr
1 molal aq. solution means 1 mole of solute in 1 kg (1000 g) of the solvent water. The number of moles of water is 1000/18 = 55.55 moles
Thus the mole fraction of the solute = Xsolute = 1 / (1 + 55.55) = 0.0177
We have, (p0 - pS) / p0 = Relative lowering in vapour pressure = Xsolute = 0.0177
Or p0 - pS = p0 . Xsolute = 760 X 0.0177 = 13.4 torr
7. Mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene (p0 = 120 torr) and toluene (p0 = 80 torr) having 2 mole each, is:
A Solution having equal moles of liquids will have equal mole fractions of the components. In the similar way, a binary solution will have the mole fraction of each component 1/2.
Here we should take in account the mole fraction both in liquid and vapour phase.
We know, p (toluene) = p0 (toluene) . X (toluene liquid) = P(Total) . X(toluene vapour) ..... equation 1
Since P (Total) = p0 (toluene) . X (toluene liquid) + p0(Benzene) . X(Benzene liquid)
Putting this into equation 1,
Hence 80 . 1/2 = {(80 X 1/2) + (120 X 1/2)} X(toluene vapour)
Thus X(toluene vapour) = 0.4
8. 25 ml of an aqueous solution of KCl solution was found to require 20 ml of 1M AgNO3 solution when titrated (K2Cr2O7 being the indicator). The depression in freezing point of KCl solution with 100% ionisation is [kf = 2.0 K kg per mol and molarity = molality]:
We know milimoles = Molarity X volume in ml
In this case eqal milimoles of AgNO3 and KCl reacts
That means milimoles of AgNO3 = Milimoles of KCl
milimoles of KCl = 20 X 1 = 20 milimoles = 20 /1000 = 0.02 moles in 25 ml of AgNO3 solution
Moles of KCl in 1000 ml (1 ltr) = 0.02 X 40 = 0.8 mole per litre
Or molarity = 0.8 = Molality
Taking abnormal colligative properties into consideration, Depression in freezing point = i . Kf . m' = 2 X 2 X 0.8 = 3.2 K
Alternatively,
Moles of AgNO3 = Moles of KCl
Since we have 20 ml of 1 molar solution, we have taken out 20 ml of an 1 molar solution in which 1 mole is dissolved in 1000 ml solvent.
Thus 20 ml contains moles = 1 / 50 = 0.02 moles = moles of KCl
Thus 25 ml of KCl solution contains 0.02 moles of KCl
or 1000 ml will contains = 0.02 X 40 = 0.8 moles
but KCl is 100 % ionised
Since 1 mole of KCl gives 2 moles, it means 0.8 moles will produce 1.6 moles of particles
Or Molarity = 1.6 = molality (given)
Hence dipression in feezing pont = Kf . m' = 2 X 1.6 3.2 K
9. If Solution A contains 6g urea in 100 ml solution and Solution B contains 6g acetic acid in 100 ml solution then they are:
The molecular mass of urea and acetic are same = 60
Both of them are dissolved in 100 ml solution, thus the molarity of both the solution are same.
but acetic acid dissociates where as urea does not. The number of particles in both the solution are not eqal. Hence they are not isotinic.
10. If you have 100 ml of 0.1 m aq. KCl solution then what should you do to make it 0.2 m?
We want to make a concentrated solution. Amount of solute in the solution must increase. We can do it by evaporating solvent or by adding solute to the given solution.
In the final 0.2m aq solution, the number of moles dissolved per 1000 gram (1000 ml for water) of solvent is 0.2
Thus 100 ml or g must contain 0.02 moles of solute in the final solution
But the given solution is 100 ml of 0.1m solution means it contains 0.01 moles of solute.
Hence we must add solute to 100 ml 0.1 m solution = 0.2 - 0.1 = 0.1 mole of solute
Topic: Chemical Kinetics (NDA, ICAR, OUAT Level)
1. In the integratd equation for the rate constant, if the slope and y - intercept are (-k/2.303) and log a (Base 10), then the equation indicates ______ order.
Looking into the integrated rate equation of the first order, we can find the one which is similar to the slope form of straight line eqation. The answer is First order.
2. On increasing the intial concentration of the reactant by three times the rate of reaction is increased by a factor 27. The order of the reaction is ____.
If initial rate = r1 = K [A]^x .......... eq. 1, where x = order of the reaction
Then as per the equation, final rate = r2 = K[3A]^x ............. eq. 2,
Thus r2/r1 = 27 = 3^x
Thus x = order of the reaction = 3
3. The reaction of the type , A+B --------> C+D, has the follwing energy diagram, The enthalpy change and the activaion energy for the reaction C+D --------> A+B are respectively ________.
The difference between the energy of the product and reactants is called the enthalpy of the reaction. For the backward reaction, when C+D give product, they have to rise to the threshold energy, which means the activation energy of C+D is y+X.
4. The rate of gaseous reaction of the type , A+B --------> product, is given by,
dx/dt = K [A] [B]. If the volume of the vessel is reduced to one fourth of initial volume, then the ratio of final rate to the initial rate is _______.
With respect to A and B the reaction is of second order. On reducing the volume by 1/4th, the concentration of both the reactants increase by 4 times because volume and concentration are inversly proportional.
r1 = K [A] [B] and r2 = K [4A] [4B]
Thus r2/r1 = 16:1
5. The decomposition of acetaldehyde is given by the following balanced reaction:
CH3CHO (g) ---------------> CH4 (g) + CO (g). The rate of reaction w.r.t. pressure of the reactant is _________.
The rate of reaction with respect to concentration of reactant is written as (-dC/dt)
We have PV = nRT
Or P = (n/v) RT
Or P = C RT (since C = concentration = n/v)
Or dP/dt = dC/dt (RT)
Hence dC/dt = (1/RT) dP/dt and (-dC/dt) = (-1/RT) dP/dt
6. Rate of formationm of SO3 in the following reaction, 2SO2 + O2 ----------> 2SO3 is 100 kg per minute. The rate of disappearance of SO2 is ______.
Its a very interesting and tricky question.
Reactants actually react in terms of their moles not in gram.
Hence we must convert the kg (kilogram) per minuite into kilo mole per minute or even in mole per minute, then we can determine the rate of disappearnce of SO2.
Given that d[SO3]/dt = 100 kilo gram per minute
Molecular mass of SO3 = 80
Thus d[SO3]/dt = 100 kilo gram per minute = 100/80 kilo mole per minute
From the reaction, d[SO3]/dt = 2/2 {-d[SO2]/dt} = -d[SO2]/dt = 100/80 kilo mole per minute
Hence, -d[SO2]/dt = 100/80 kilo mole per minute = (100/80) X 64 kilo gram per minute
(since mole X molar mass = given mass)
7. The ppoint A in the graph may represent to __________.
The graph may represent an SN1 reaction in which an intermediate is formed. The point A represent a comparatively lower energetic states means a comparatively stable state. This indicates that we should choose from the options " may be an intermediate like a tertiary carbocation which is a stable intermediate".
8. A first order reaction is complete in 10 mimutes. Calculate the time required for the completion of 90% of the reaction.
t1 = k/2.303 log a/(a-x) = k/2.303 log 100/(100-10) = k/2.303 log 100/90 = 10 minutes
t2 = k/2.303 log 100/(100-90) =k/2.303 log 100/10
Take ratio of these two equations and solve to get the value of t2.
9. 
Comparing with the logarithimic expression of Arrhenius equation and the given equation,
log k = log A - Ea/2.303RT and log K = 14 - (1.25 X 10^4 K) / T
Thus log A = 14
=> The pre -exponential factor = 10^14
10.
From the graph it is clear that the half life period remains constant whatever the initial concentration is. This trend is found in the first order reaction.
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